Question

Reflection of the line $\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z-4}{1}$ in the plane $x+y+z$ is:

• A. $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-4}{1}$
• B. $\frac{x-1}{-3}=\frac{y-2}{-1}=\frac{z-4}{1}$
• C. $\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$
• D. $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-4}{1}$

Answer 1

The correct option is A $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-4}{1}$

$\begin{array}{c}\text{Solution - Given line}\equiv \frac{x-1}{-1}=\frac{y-2}{3}=\frac{2-4}{1}=\lambda \\ \text{and plane:}x+y+z=0\\ (-1)\left(1\right)+3x|+|x\mid =3\Rightarrow \text{they do not intersect}\\ \text{let}P\text{be a point on the line}\\ \therefore P=(1-\lambda ,3\lambda +2,\lambda +4)\\ \text{P satisfies the equation of plane}\\ \therefore 1-\lambda +3\lambda +2+\lambda +4=0\\ 3\lambda =-7\\ \lambda =-\frac{7}{3}\end{array}$

$\begin{array}{c}P(\frac{10}{3},-5,\frac{5}{3})\\ \text{Let}M(x,y,z)\\ \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}=-\frac{2(10+2+3)}{3}\\ \therefore \text{line comes out to be}\\ \frac{x-1}{3}=\frac{y-2}{1}=\frac{z-4}{1}\\ \text{Hence,}\left(A\right)\text{is the correct option}\end{array}$